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Fed [463]
3 years ago
12

Is y=6/x-2 linear or nonlinear

Mathematics
2 answers:
zhannawk [14.2K]3 years ago
8 0

Answer:

Non linear. It is a curve not a line.

Vikki [24]3 years ago
4 0
No liner
Reason? Because 8 is a negative number
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A student said that 42% of 78 is 45. Is this answer reasonable? Explain.
I am Lyosha [343]
42% = 0.42   
0.42 * 78 = 32.76
7 0
3 years ago
Read 2 more answers
4. What number is its own additive inverse? Why?​
exis [7]

Answer:

0 (zero)

Step-by-step explanation:

It is the only Rational number, which is its own additive inverse.

Additive inverse of ….

2 is -2

-3 is 3

7/9 is -7/9

-1 is 1

1 is -1

I gave a few examples above, each is additive inverse of the other because their sum = 0. For finding additive inverse we only need to change its sign., and if we need both numbers equal, it has to be 0. Since 0+0 = 0.

<em>Hope this Helps!!!</em>

5 0
3 years ago
Please help me out........................
grandymaker [24]

RM is 10 because it’s the same as PM

Hope this helps

7 0
4 years ago
(03.01 LC) Given that h(x) = 3x −19, find the value of x that makes h(x) = 71. (5 points)
LekaFEV [45]
The function given is:
h(x) = 3x-19
we want to find the value of x that makes this function equal to 71. To do so, we will substitute h(x) with 71 and solve for x as follows:
h(x) = 3x - 19
71 = 3x - 19
3x = 71+19
3x = 90
x = 90/3 = 30
6 0
3 years ago
Read 2 more answers
Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.
skad [1K]

Answer:

The absolute maximum and minimum is 20\; \text{and} -20.

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

f_x=y-4=0

f_y=x=0

The only solution to this system of equations is the point (0, 4), which lies in the domain.

f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1

\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1

\therefore (0,4) is a saddle point.

Boundary points -  (4,0),  (-4,0), (0,16)

Along boundary  y=16-x^2

   f=x(16-x^2)-4x

=16x-x^3-4x

\Rightarrow f^'=16-3x^2-4=0

\Rightarrow 3x^2=12

\Rightarrow x=\pm2,\;\;y=14

Values of f(x) at these points.

\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}

Therefore, the absolute maximum and minimum is 20\; \text{and} -20.

6 0
4 years ago
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